Arithmetic Mean

$\text{Let the three numbers be x,y and z}\\ a=\frac{x+y+z}{3}\\ b=\frac{x^{2}+y^{2}+z^{2}}{3}\\ \text{Let the A.M. pair-wise product be}\ \alpha\\ \alpha=\frac{xy+yz+xz}{3}\\ \text{We know that,}\\ xy+yz+xz=\frac{(x+y+z)^{2}-(x^{2}+y^{2}+z^{2})}{2}\\ \implies\ \alpha=\frac{(x+y+z)^{2}-(x^{2}+y^{2}+z^{2})}{6}\\ \alpha=\frac{9a^{2}-3b}{6}\\ \alpha=\boxed{\frac{3a^{2}-b}{2}}$

Let the three numbers be $x$ , $y$ and $z$ , then $a = \dfrac {x+y+z}3$ , $b = \dfrac {x^2+y^2+z^2}3$ and the pairwise arithmetic mean is $\dfrac {xy+yz+zx}3$ . We know that:

$\begin{aligned} (x+y+z)^2 & = x^2+y^2+z^2 + 2(xy+yz+zx) \\ \implies 2(xy+yz+zx) & = (x+y+z)^2 - x^2+y^2+z^2 \\ & = (3a)^2 - 3b \\ \implies \frac {xy+yz+zx}3 & = \frac {(3a)^2 - 3b}6 \\ & = \boxed{\dfrac {3a^2-b}2} \end{aligned}$